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This is Exercise 3.3.7 of Robinson's "A Course in the Theory of Groups (Second Edition)." According to Approach0, it is new to MSE.

The Details:

On page 28, ibid.,

A right operator group is a triple $(G, \Omega, \alpha)$ consisting of a group $G$, a set $\Omega$ called the operator domain and a function $\alpha:G\times \Omega\to G$ such that $g\mapsto (g,\omega)\alpha$ is an endomorphism of $G$ for each $\omega\in\Omega$. We shall write $g^\omega$ for $(g,\omega)\alpha$ and speak of the $\Omega$-group if the function $\alpha$ is understood.

[. . .]

If $G$ is an $\Omega$-group, an $\Omega$-subgroup of $G$ is a subgroup $H$ which is $\Omega$-admissible, that is, such that $h^\omega\in H$ whenever $h\in H$ and $\omega\in\Omega$.

On page 60, ibid.,

[Consider] a partially ordered set $\Lambda$ with partial order $\le$. We say that $\Lambda$ satisfies the maximal condition of each nonempty subset $\Lambda_0$ contains at least one maximal element.

Analogously, the minimal condition is defined.

On page 80, ibid.,

Let $G$ be a group with operator domain $\Omega$. An $\Omega$-subgroup $H$ is called an $\Omega$-direct factor of $G$ If there exists an $\Omega$-subgroup $K$ such that $G=H\times K$; then $K$ is called an $\Omega$-direct complement of $H$ in $G$. If there are no proper nontrivial $\Omega$-direct factors of $G$, then $G$ is said to be $\Omega$-indecomposable [. . .].

On page 81, ibid

[The maximal and minimal conditions on $\Omega$-direct factors being equivalent guarantees that an] $\Omega$-group may be expressed as a direct product of finitely many nontrivial $\Omega$-indecomposable subgroups: such a direct decomposition is called a Remak decomposition.

Theorem (Remak): If an $\Omega$-group $G$ has the minimal condition on $\Omega$-direct factors, it has a Remak decomposition.

On page 92, ibid., there are the following two exercises:

Exercise 3.3.4: Prove that the central automorphisms of a group $G$ form a subgroup ${\rm Aut}_{{\rm c}}\ G$ of ${\rm Aut}\ G$.

Proof: Simply apply the one-step subgroup test.$\square$

Exercise 3.3.5 (J.E. Adney and Ti Yen): Let $G$ be a finite group which has no nontrivial abelian direct factors. Prove that $\lvert{\rm Aut}_{{\rm c}}\ G\rvert =\lvert {\rm Hom}(G_{{\rm ab}}, Z(G))\rvert$. Deduce that if $G$ has no nontrivial central automorphisms, then $Z(G)\le G'$.

Proof: This follows rather easily from the hint: "If $\alpha\in {\rm Aut}_{{\rm c}}\ G$, define $\theta_\alpha\in{\rm Hom}(G_{{\rm ab}}, Z(G))$ by $(gG')^{\theta_\alpha}=g^{-1}g^\alpha$. Show that $\alpha\mapsto \theta_\alpha$ is a bijection."$\square$

The Question:

If $G$ is a finite group and $|G_{{\rm ab}}|$ and $|Z(G)|$ are coprime, prove that $G$ has a unique Remak decomposition.

Thoughts:

I think that Exercise 3.3.5 (along with, say, Lagrange's theorem) is required but I'm not sure how to implement it here.

Of course, there is, on page 83, the

Theorem (Krull, Remak, Schmidt): Let $G$ be an $\Omega$-group satisfying the maximal and minimal conditions on normal $\Omega$-subgroups. If

$$G=H_1\times \dots\times H_r=K_1\times \dots \times K_s$$

are two Remak decompositions of $G$, then $r=s$ and there is a central $\Omega$-automorphism $\alpha$ of $G$ such that, after suitable relabelling of the $K_j$s if necessary, $H_i^\alpha=K_i$ and $$G=K_1\times \dots \times K_k\times H_{k+1}\times \dots \times H_r$$ for $k=1,\dots, r$.

It is noted, though, shortly after the proof, that this theorem does not guarantee the uniqueness of the Remak decomposition.

There is a couple of other, rather technical theorems on uniqueness of Remak decompositions in the ensuing pages but I shall not share them here, since this is too long a post already and I don't see how to apply them.

This is not an exercise I think I can do myself any time soon.

I am reminded of the Fundamental Theorem of Arithmetic, say, or the Fundamental Theorem of Finitely Generated Abelian Groups; so these are similar problems I am familiar with, although the latter is not yet proven in Robinson's book.

The kind of answer I'm hoping for is a full solution. This is a big ask, so I'll be happy with strong hints or perhaps something like a list of bullet points of intermediate steps.


Please help :)

Shaun
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    I think this is easier than you think. You know from Exercise 3.3.5 and the given coprimeness condition that ${\rm Aut_c}(G)$ is trivial. Now apply the Krull Remak Schmidt Theorem with $k=r$ and you find that the two decompositions are the same after reordering the factors. – Derek Holt Jun 16 '21 at 17:04
  • Thank you, @DerekHolt. One thing I don't understand is how the no nontrivial abelian direct factors part of that exercise satisfied. Please would you write up an answer? – Shaun Jun 16 '21 at 18:38
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    An abelian direct factor would contradict the assumed coprimeness condition - think about it! – Derek Holt Jun 16 '21 at 18:53

1 Answers1

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Here is a brief answer. We are told that $\gcd(|Z(G)|,|G_{\rm ab}|)=1$. But we know from Exercise 3.3.5 that $|{\rm Aut_c}(G)| = |{\rm Hom}(G_{\rm ab},Z(G))|$, so $|{\rm Aut_c}(G)|=1$.

Now, if $G$ has two Remak decompositions then we can apply the Krull Remak Schmidt Theorem with $k=r$, and the elements $\alpha \in {\rm Aut_c}(G)$ must be equal to the identity, so the two decompositions are the same after permuting the direct factors.

If one of the factors, say $H_1$ was nontrivial and abelian, then we would have $H_1 \le Z(G)$, but $G/(H_2 \times \cdots H_r) \cong H_1$, so $H_1$ is isomorphic to a quotient group of $G_{\rm ab}$, and hence $|H_1|$ divides $\gcd(|Z(G)|,|G_{\rm ab}|)=1$, contradiction.

Shaun
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Derek Holt
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